Tomorrow about yarn for $39 and two pairs of knitting needles her total cost for yarn and eating noodles and $69 tomorrow right in equation that represents the cost of each pair of knitting needles OK represents the cost of each pair of knitting needles which statement about verifying the value of K are true check all that apply

Statement A

Statement C

Step-by-step explanation:

Tamara bought yarn for $39 and 2 pairs of knitting needles. Her total cost for the yarn and knitting needles was $69. Tamara writes an equation that represents the cost of each pair of knitting needles. If k represents the cost of each pair of knitting needles, which statements about verifying the value of k are true? Check all that apply.

A. Tamara should substitute into the equation . After she substitutes the correct value for k into the equation, Tamara should find that the left side of the equation is equal to the right side

69 = 39 + 2k

69 = 39 + 2(15)

69 = 39 + 30

69 = 69

B. Tamara should substitute into the equation . After she substitutes the correct value for k into the equation, Tamara should find that the left side of the equation is not equal to the right side.

C. To verify the solution to the equation, Tamara multiplies the solution by 2 and then adds it to 39. Then, she checks that it is equal to 69.

2 × 15 = 30

39 + 30 = 69

D. To verify the solution to the equation, Tamara adds the solution to 2 and then adds it to 69. Then, she checks that it is equal to 39.

Cost of Yarn = $39

2 Knitting needles = $2k

Total cost = cost of yarn + cost of 2 knitting needles

69 = 39 + 2k

69 - 39 = 2k

30 = 2k

k = 30/2

= 15

k = $15

Statement A and statement C are true

answer: .75 km

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)